"""
You are given a 0-indexed string s that has lowercase English letters in its even indices and digits in its odd indices.

There is a function shift(c, x), where c is a character and x is a digit, that returns the xth character after c.

For example, shift('a', 5) = 'f' and shift('x', 0) = 'x'.
For every odd index i, you want to replace the digit s[i] with shift(s[i-1], s[i]).

Return s after replacing all digits. It is guaranteed that shift(s[i-1], s[i]) will never exceed 'z'.



Example 1:

Input: s = "a1c1e1"
Output: "abcdef"
Explanation: The digits are replaced as follows:
- s[1] -> shift('a',1) = 'b'
- s[3] -> shift('c',1) = 'd'
- s[5] -> shift('e',1) = 'f'
Example 2:

Input: s = "a1b2c3d4e"
Output: "abbdcfdhe"
Explanation: The digits are replaced as follows:
- s[1] -> shift('a',1) = 'b'
- s[3] -> shift('b',2) = 'd'
- s[5] -> shift('c',3) = 'f'
- s[7] -> shift('d',4) = 'h'


Constraints:

1 <= s.length <= 100
s consists only of lowercase English letters and digits.
shift(s[i-1], s[i]) <= 'z' for all odd indices i.

"""

import re


class Solution(object):
    @staticmethod
    def callback(match):
        next_char_code_incr = int(match.group(2))
        return match.group(1) + chr(ord(match.group(1)) + next_char_code_incr)

    def replaceDigits(self, s):
        """
        :type s: str
        :rtype: str
        """
        return re.sub(r'(?:([a-z])([0-9]))', self.callback, s)
